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$\left[ \frac{1}{2}+\frac{1}{\pi}\tan^{-1}(\lambda_n)\right]^n\rightarrow 1$.

I want to find the condition on $\lambda_n$ which leads the above display as $n$ tends to $\infty$.

I guess that the condition would be $\lambda_n \succ n$, but I cannot show it in a rigorous way.

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  • $\begingroup$ Could you explain what connection this question has with statistics? $\endgroup$ – whuber Nov 12 '15 at 0:04
  • $\begingroup$ This comes from a statistical problem. For example, one can show that $P\{\sum_{i=1}^n \exp( W_i/2 ) >n \lambda_n \}\leq 1- [P(Q \leq \lambda)]^n$, where $W_i$ is from i.i.d $\chi^2_1$ for $i=1,\cdots,n$ and $Q$ follows Cauchy distribution. Also, $[P(Q \leq \lambda)]^n = [1/2+\tan^{-1}(\lambda_n)/\pi]^n$. When $W_i$ is a test statistic, it is related to a multiple testing problem. $\endgroup$ – Minsuk Shin Nov 12 '15 at 2:19
  • $\begingroup$ It's a typo; $P(Q\leq \lambda)$ should be $P(Q\leq \lambda_n)$. $\endgroup$ – Minsuk Shin Nov 12 '15 at 2:31
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If $\left[X_n\right]^n \rightarrow 1$ as $n \rightarrow \infty$, then $X_n \rightarrow 1$ as $n \rightarrow \infty$.

It follows that $\left[\frac{1}{2} + \frac{1}{\pi}\tan^{-1}\left(\lambda_n\right)\right]$ must tend to 1 as $n$ tends to $\infty$.

Therefore, $\frac{1}{\pi}\tan^{-1}\left(\lambda_n\right) \rightarrow \frac{1}{2}$, which implies that $ \tan^{-1}\left(\lambda_n\right) \rightarrow \frac{\pi}{2}$.

Finally, this requires that as $n \rightarrow \infty$, $\lambda_n$ must tend towards $-\tan\left(\frac{\pi}{2}\right)$

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    $\begingroup$ Although necessary, that is not a sufficient condition. $\endgroup$ – whuber Nov 12 '15 at 1:15
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    $\begingroup$ Thankyou for the comments. This is obvious in a sense that $\tan(\pi/2)=\infty$ and implicitly it is assumed that $\lambda_n \rightarrow \infty$. I am interested the rate of $\lambda$. $\endgroup$ – Minsuk Shin Nov 12 '15 at 2:29

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